Let A (sec$$\theta$$, 2tan$$\theta$$) and B (sec$$\phi$$, 2tan$$\phi$$), where $$\theta$$ + $$\phi$$ = $$\pi$$/2, be two points on the hyperbola 2x2 $$-$$ y2 = 2. If ($$\alpha$$, $$\beta$$) is the point of the intersection of the normals to the hyperbola at A and B, then (2$$\beta$$)2 is equal to ____________.
Your Input ________
Answer
Correct Answer is Bonus
Explanation
Since, point A (sec$$\theta$$, 2tan$$\theta$$) lies on the hyperbola 2x2 $$-$$ y2 = 2
but according to question $$\theta$$ + $$\phi$$ = $${\pi \over 2}$$ which is not possible.
Hence, it must be a 'BONUS'.
2
JEE Main 2021 (Online) 27th August Morning Shift
Numerical
If the minimum area of the triangle formed by a tangent to the ellipse $${{{x^2}} \over {{b^2}}} + {{{y^2}} \over {4{a^2}}} = 1$$ and the co-ordinate axis is kab, then k is equal to _______________.
So, area $$(\Delta OAB) = {1 \over 2} \times {b \over {\cos \theta }} \times {{2a} \over {\sin \theta }}$$
$$ = {{2ab} \over {\sin 2\theta }} \ge 2ab$$
$$\Rightarrow$$ k = 2
3
JEE Main 2021 (Online) 27th July Evening Shift
Numerical
Let E be an ellipse whose axes are parallel to the co-ordinates axes, having its center at (3, $$-$$4), one focus at (4, $$-$$4) and one vertex at (5, $$-$$4). If mx $$-$$ y = 4, m > 0 is a tangent to the ellipse E, then the value of 5m2 is equal to _____________.
If the point on the curve y2 = 6x, nearest to the point $$\left( {3,{3 \over 2}} \right)$$ is ($$\alpha$$, $$\beta$$), then 2($$\alpha$$ + $$\beta$$) is equal to _____________.
Your Input ________
Answer
Correct Answer is 9
Explanation
Let, $$P \equiv \left( {{3 \over 2}{t^2},3t} \right)$$ is on the curve.